Problem: Solve for $x$ : $3x^2 - 6x + 3 = 0$
Solution: Dividing both sides by $3$ gives: $ x^2 {-2}x + {1} = 0 $ The coefficient on the $x$ term is $-2$ and the constant term is $1$ , so we need to find two numbers that add up to $-2$ and multiply to $1$ The number $-1$ used twice satisfies both conditions: $ {-1} + {-1} = {-2} $ $ {-1} \times {-1} = {1} $ So $(x - {1})^2 = 0$ $x - 1 = 0$ Thus, $x = 1$ is the solution.